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\title{Oefeningen TAI: \\
Oefenzitting 1: Samenstellingswetten en groepen}
\date{}
\maketitle





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% About
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\section*{About}
Uitwerking van oefenziting 1 van Toepassingen van Algebra in Informatica gedoceerd in de 3e Bachelor Informatica aan de KULeuven in 2012.

Latex code van dit document te vinden op:\\
SVN checkout: \texttt{https://oefenzittingen-tai.googlecode.com/svn/trunk/}\\
Google code: \texttt{https://code.google.com/p/oefenzittingen-tai/}

Credits: Peter Roelants, Wouter Schaekers, Daan Wendelen
%TODO: put your name here

Te gebruiken op eigen risico!



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% Oef 1 
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\section*{Opgave 1}
Op $\mathbb{R}$ defini\"eren we de samenstellingswet $\alpha \tau \beta = \alpha + \beta + \alpha^2 \beta^2$.

(a) Deze wet heeft een neutraal element. Welk?\\
(b) Ze is niet associatief. Ga na!\\
(c) Ze is commutatief. Waarom?\\


\subsection*{oplossing 1}
Samenstellingswetten: p. 79 eerste deel cursus.

(a) $0$, want $\forall r : r \tau 0 = r = 0 \tau r$\\
(b) $(1 \tau 1) \tau 2 = 41 \neq 57 = 1 \tau (1 \tau 2)$\\
(c) Ok, want $+$ en $\cdot$ zijn commutatief in $\mathbb{R}$, \\
$\alpha \tau \beta = \alpha + \beta + \alpha^2 \beta^2 = \beta + \alpha + \beta^2 \alpha^2 = \beta \tau \alpha$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Oef 2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Opgave 2}
Bewijs dat in $\mathbb{R}^2 \times \mathbb{R}^2$ volgende relaties equivalentierelaties zijn:

\begin{center}
$G = \{ ((a,b),(c,d) | a^2 + b^2 = c^2 + d^2 \}$\\
$H = \{ ((a,b),(c,d) | b - a = d - c \}$\\
$J = \{ ((a,b),(c,d) | a + b = c + d \}$\\
\end{center}

Welke zijn de partities die hierdoor gedefinieerd worden? Welke partitie definieert $H \cap J$?


\subsection*{oplossing 2}
Definitie equivalentierelatie: p. 60 eerste deel cursus
Definitie partitie: p.61 eerste deel cursus

($G$): Partitie: $G_{(a,b)} = \{ (x,y) \in \mathbb{R}^2 | (x - 0)^2 + (y - 0)^2 = \sqrt{a^2 + b^2}^2 \}$ $\rightarrow$ dit komt overeen met de cirkel met middelpunt $(0,0)$ en straal $\sqrt{a^2 + b^2}$.

($H$): Partitie: $H_{(a,b)} = \{ (x,y) \in \mathbb{R}^2 | y = x + (b - a) \} $

($J$): Partitie: $J_{(a,b)} = \{ (x,y) \in \mathbb{R}^2 | y = -x + (b + a) \} $

($H \cap J$): Partitie: ${H \cap J}_{(a,b)} = \{ (x,y) \in \mathbb{R}^2 | x = a\ en\ y = b \} = \{ (a,b) \} $

Aangezien de bovenstaande relaties elk een partitie defini\"eren zijn ze ook equivalentierelaties. Partie en equivalentierelatie zijn namelijk isomorfe begrippen (p. 62 eerste deel cursus).



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% Oef 3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Opgave 3}
Los het volgende stelsel op:
\[ 
\begin{cases} 
3x_1 - 2x_2 + 6x_3 = 4 \pmod 7 \\ 
4x_1 + x_2 + x_3 = 0 \pmod 7 \\
2x_1 + x_2 + 2x_3 = -1 \pmod 7 \\ 
\end{cases} 
\]


\subsection*{oplossing 3}
Stelsel oplossen zoals normaal en alle co\"effici\"enten modulo 7 doen:

$x_1 = 2, \\
x_2 = 3, \\
x_3 = 3$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Oef 4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Opgave 4}
Bepaal de isometri\"en van een gelijkzijdige driehoek. Stel voor deze
isometrien de bewerkingstabel op, onder de samenstellingwet $\circ$.



\subsection*{oplossing 4}
Bewerkingstabel: p.80 eerste deel cursus.


% r1
\begin{tikzpicture} 
\coordinate [label=left:{$e$: identiteit:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
\coordinate [] (r_ar) at (3,0);
\draw[->] (l_ar) -- (r_ar);

\coordinate [label=above:{$A$}] (A2) at (3.9,0.3); 
\coordinate [label=left:{$C$}] (C2) at (3.6,-0.3); 
\coordinate [label=right:{$B$}] (B2) at (4.2,-0.3);
\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}

% r1
\begin{tikzpicture} 
\coordinate [label=left:{$r_1$: rotatie $120 \degree$ CW:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
\coordinate [] (r_ar) at (3,0);
\draw[->] (l_ar) -- (r_ar);

\coordinate [label=above:{$C$}] (C2) at (3.9,0.3); 
\coordinate [label=left:{$B$}] (B2) at (3.6,-0.3); 
\coordinate [label=right:{$A$}] (A2) at (4.2,-0.3);
\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}

% r2
\begin{tikzpicture} 
\coordinate [label=left:{$r_2$: rotatie $120 \degree$ CCW:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
\coordinate [] (r_ar) at (3,0);
\draw[->] (l_ar) -- (r_ar);

\coordinate [label=above:{$B$}] (B2) at (3.9,0.3); 
\coordinate [label=left:{$A$}] (A2) at (3.6,-0.3); 
\coordinate [label=right:{$C$}] (C2) at (4.2,-0.3);
\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}

% s1
\begin{tikzpicture} 
\coordinate [label=left:{$s_1$: spiegeling C-B:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
\coordinate [] (r_ar) at (3,0);
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\coordinate [label=above:{$A$}] (A2) at (3.9,0.3); 
\coordinate [label=left:{$B$}] (B2) at (3.6,-0.3); 
\coordinate [label=right:{$C$}] (C2) at (4.2,-0.3);
\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}

% s2
\begin{tikzpicture} 
\coordinate [label=left:{$s_2$: spiegeling C-A:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
\coordinate [] (r_ar) at (3,0);
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\coordinate [label=above:{$C$}] (C2) at (3.9,0.3); 
\coordinate [label=left:{$A$}] (A2) at (3.6,-0.3); 
\coordinate [label=right:{$B$}] (B2) at (4.2,-0.3);
\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}

% s3
\begin{tikzpicture} 
\coordinate [label=left:{$s_3$: spiegeling A-B:}] (text) at (0,0);

\coordinate [label=above:{$A$}] (A1) at (1.3,0.3); 
\coordinate [label=left:{$C$}] (C1) at (1,-0.3); 
\coordinate [label=right:{$B$}] (B1) at (1.6,-0.3);

\draw (A1) -- (B1); 
\draw (A1) -- (C1);
\draw (C1) -- (B1);

\coordinate [] (l_ar) at (2.2,0);
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\draw (A2) -- (B2); 
\draw (A2) -- (C2);
\draw (C2) -- (B2);
\end{tikzpicture}


%tabel
\begin{table}[h!]
\centering
\begin{tabular}{ | c | c c c c c c | }
	\hline
	$\circ$	&	$e$ &	$r_1$ &	$r_2$ &	$s_1$ &	$s_2$ &	$s_3$\\
  	\hline
    $e$		&	$e$ &	$r_1$ &	$r_2$ &	$s_1$ &	$s_2$ &	$s_3$\\           
	$r_1$	&	$r_1$ &	$r_2$ &	$e$ &	$s_3$ &	$s_1$ &	$s_2$\\
	$r_2$	&	$r_2$ &	$e$ &	$r_1$ &	$s_2$ &	$s_3$ &	$s_1$\\
	$s_1$	&	$s_1$ &	$s_2$ &	$s_3$ &	$e$ &	$r_1$ &	$r_2$\\
	$s_2$	&	$s_2$ &	$s_3$ &	$s_1$ &	$r_1$ &	$e$ &	$r_1$\\
	$s_3$	&	$s_3$ &	$s_1$ &	$s_2$ &	$r_1$ &	$r_2$ &	$e$\\
  	\hline  
\end{tabular}
\end{table}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Oef 5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Opgave 5}
Een latijns vierkant is een $n \times n$ tabel waarin slechts $n$ verschillende elementen voorkomen. In elke rij en elke kolom komt namelijk elk element juist eenmaal voor.

(a) Bewijs dat de bewerkingstabel voor een eindige groep steeds een Latijns vierkant is.

(b) Is dit ook een voldoende voorwaarde om een groep te hebben? Bepaal of volgend Latijns vierkant de bewerkingstabel van een groep is:

%tabel
\begin{table}[h!]
\centering
\begin{tabular}{ | c | c c c c c c | }
	\hline
	$\tau$	&	$a$ &	$b$ &	$c$ &	$d$ &	$e$ &	$f$\\
  	\hline
    $a$		&	$c$ &	$e$ &	$a$ &	$b$ &	$f$ &	$d$\\           
	$b$		&	$f$ &	$c$ &	$b$ &	$a$ &	$d$ &	$e$\\
	$c$		&	$a$ &	$b$ &	$c$ &	$d$ &	$e$ &	$f$\\
	$d$		&	$e$ &	$a$ &	$d$ &	$f$ &	$c$ &	$b$\\
	$e$		&	$d$ &	$f$ &	$e$ &	$c$ &	$b$ &	$a$\\
	$f$		&	$b$ &	$d$ &	$f$ &	$e$ &	$a$ &	$c$\\
  	\hline  
\end{tabular}
\end{table}


\subsection*{oplossing 5}
Definitie groep: p. 93 eerste deel cursus, p. 1 tweede deel cursus.

(a) eindig aantal elementen $\rightarrow$ closure\\
1 element per rij $\rightarrow$ alle elementen zijn regulier

(b) closure: $\tau$ is overal bepaald (eindig aantal elementen, heel vierkant is opgevuld)\\
neutraal: $c$ is het neutraal element\\
inverse: $\tau$ is symmetriseerbaard omdat elk element een inverse heeft door de vorm van het vierkant\\
associatief: $\tau$ is niet associatief: $a \tau (a \tau b) = f \neq e = (a \tau a) \tau b$\\

Het is dus GEEN voldoende voorwaarde dat de bewerkingstabel een Latijns vierkant is om een groep te hebben.

\end{document}